Step-by-step explanation:
If you like my answer than please mark me brainliest and follow me thanks
That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:
-- 0:
A number greater than 10 with a 0 in the units place is a multiple of
either 5 or 10, so it's not a prime number.
-- 1:
A number greater than 10 with a 1 in the units place could be
a prime (11, 31 etc.) but it doesn't have to be (21, 51).
-- 2:
A number greater than 10 with a 2 in the units place has 2 as a factor
(it's an even number), so it's not a prime number.
-- 3:
A number greater than 10 with a 3 in the units place could be
a prime (13, 23 etc.) but it doesn't have to be (33, 63) .
-- 4:
A number greater than 10 with a 4 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 5:
A number greater than 10 with a 5 in the units place is a multiple
of either 5 or 10, so it's not a prime number.
-- 6:
A number greater than 10 with a 6 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 7:
A number greater than 10 with a 7 in the units place could be
a prime (17, 37 etc.) but it doesn't have to be (27, 57) .
-- 8:
A number greater than 10 with a 8 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 9:
A number greater than 10 with a 9 in the units place could be
a prime (19, 29 etc.) but it doesn't have to be (39, 69) .
So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.
It CAN't have a 0, 2, 4, 5, 6, or 8 .
The only choice that includes all of the possibilities is 'A' .
Answer:
0.625
Step-by-step explanation:
Given that {A1, A2} be a partition of a sample space and let B be any event. State and prove the Law of Total Probability as it applies to the partition {A1, A2} and the event B.
Since A1 and A2 are mutually exclusive and exhaustive, we can say
b) P(B) = P(A1B)+P(A2B)
Selecting any one coin is having probability 0.50. and A1, A2 are events that the coins show heads.
c) Using Bayes theorem
conditional probability that it wasthe biased coin
=
d) Given that the chosen coin flips tails,the conditional probability that it was the biased coin=
Answer:
2 equations
28.8/100*75 = $21.6 - 10% 21.6/100*90 = $19.44
Step-by-step explanation:
First one we determine the 25% ans second we deduct 10% from that result.
